Question

A photoelectric material having work-function ${\varphi }_0$ is illuminated with light of wavelength $\lambda \left(\lambda <\frac{hc}{{\varphi }_0}\right)$ . The fastest photoelectron has a de Broglie wavelength ${\lambda }_d$ . A change is the wavelength of the incident light by $\Delta \lambda$ result in change $\Delta {\lambda }_{d}in{\lambda }_d$ .then the ratio $\Delta {\lambda }_d/\Delta \lambda$ is proportional to

• A. $\frac{{\lambda }_d^2}{{\lambda }^2}$
• B. $\frac{{\lambda }_d}{\lambda }$
• C. $\frac{{\lambda }_d^3}{\lambda }$
• D. $\frac{{\lambda }_d^3}{{\lambda }^2}$

Answer 1

Answer: (A)

$\frac{{\lambda }_d^2}{{\lambda }^2}$

$\frac{hc}{\lambda }-{\varphi }_0=KE=\frac{P^2}{2me}=\frac{h^2}{2me{\lambda }^2}$

Differentiating both sides

$\frac{-hc}{{\lambda }^2}d\lambda =\frac{h^2}{2me}(\frac{-2}{{\lambda }_d^3}\times d\lambda d)$

$\therefore \frac{d\lambda d}{d\lambda }\alpha \frac{\lambda d^3}{{\lambda }^2}$